Download DISCRIMINANTS AND RAMIFIED PRIMES 1. Introduction

DISCRIMINANTS AND RAMIFIED PRIMES
KEITH CONRAD
1. Introduction
A prime number p is said to be ramified in a number field K if the prime ideal factorization
(1.1)
e
(p) = pOK = pe11 · · · pgg
has some ei greater than 1. If every ei equals 1, we say p is unramified in K.
Example 1.1. In Z[i], the only prime which ramifies is 2: (2) = (1 + i)2 .
Example 1.2. Let K = Q(α) where α is a root of f (X) = T 3 − 9T − 6. Then 6 =
α3 − 9α = α(α − 3)(α + 3). For m ∈ Z, α + m has minimal polynomial f (T − m) in Q[T ],
so NK/Q (α + m) = −f (−m) = m3 − 9m + 6 and the principal ideal (α − m) has norm
N(α − m) = |m3 − 9m + 6|.
Therefore N(α) = 6, N(α − 3) = 6, and N(α + 3) = 6. It follows that (α) = p2 p3 ,
(α − 3) = p02 p3 , and (α + 3) = p02 p3 (so, in particular, α + 3 and α − 3 are unit multiples of
each other). Thus
3
(2)(3) = (6) = (α)(α − 3)(α + 3) = p2 p02
2 p3 ,
so (2) = p22 p02 and (3) = p33 . This shows 2 and 3 are ramified in K. Note that one of the
exponents in the factorization of (2) exceeds 1, while the other equals 1.
One way to think about ramified primes is in terms of the ring structure of OK /(p). By
(1.1) and the Chinese Remainder Theorem,
(1.2)
e
OK /(p) ∼
= OK /pe11 × · · · × OK /pgg .
If some ei is greater than 1, then the quotient ring OK /pei i has a nonzero nilpotent element
(use the reduction modulo pei i of any element of pi − pei i ), so the product ring (1.2) has a
nonzero nilpotent element. If each ei equals 1, then OK /(p) is a product of (finite) fields,
and a product of fields has no nonzero nilpotent elements. Thus, p ramifies in K if and only
if OK /(p) has a nonzero nilpotent element.
Our goal in this handout is to prove the following result, which characterizes the prime
numbers ramifying in a number field in terms of the discriminant.
Theorem 1.3. For a number field K, the primes which ramify are those dividing the integer
discZ (OK ).
Since discZ (OK ) 6= 0, only finitely many primes ramify in K.
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2
KEITH CONRAD
2. The power-basis case
We will first consider Theorem 1.3 when OK has a power basis over Z. The treatment of
the general case in Section 3 will not rely on the power-basis case at all, but the power-basis
case is technically simpler. We will just sketch the basic ideas behind this special case.
Proof. Assume OK = Z[α] for some α ∈ OK . Let f (T ) be the minimal polynomial of α
over Q, so f (T ) is monic in Z[T ]. With pOK factoring into primes as in (1.1), by definition
p ramifies in OK if and only if some ei > 1. The factorization of the mod p reduction f (T )
in (Z/pZ)[T ] is
e
f (T ) = π1e1 · · · · πgg
for some distinct monic irreducibles πi ∈ (Z/pZ)[T ].
The πi (T )’s are separable (all irreducibles over a finite field are separable), so some ei > 1
if and only if f (T ) has a repeated root in a splitting field over Z/pZ. This is equivalent to
f (T ) having discriminant 0: p ramifies in OK if and only if disc(f ) = 0 in Z/pZ. Since the
discriminant of a monic polynomial is a universal polynomial in its coefficients (consider
the quadratic case, where T 2 + bT + c has discriminant b2 − 4c), discriminants of monic
polynomials behave well under reduction: disc(f (T ) mod p) = disc(f (T )) mod p. Therefore
disc(f (T )) = 0 in Z/pZ if and only if disc(f ) ≡ 0 mod p. Thus p ramifies in OK if and only
if p| disc(f ). Since OK = Z[α] ∼
= Z[T ]/(f (T )), disc(f ) = discZ (Z[T ]/(f )) = discZ (OK ). 3. The general case
To prove Theorem 1.3 for every OK , we will examine discriminants of ring extensions
to show computing the discriminant commutes with reduction mod p: discZ (OK ) mod p =
discZ/pZ (OK /(p)). Then we will use (1.2) to write discZ/pZ (OK /(p)) as a product of discriminants of rings of type OK /pe and compute the discriminants of these particular rings.
Definition 3.1. Let A be a commutative ring and B be a ring extension of A which is a
finite free A-module:
B = Ae1 ⊕ · · · ⊕ Aen .
Then we set
discA (e1 , . . . , en ) = det(TrB/A (ei ej )) ∈ A.
Remark 3.2. The discriminant of a basis is an algebraic concept of “volume”. To explain
this viewpoint, we should think about TrB/A (xy) as an analogue of the dot product v · w in
Rn . For a basis v1 , . . . , vn in Rn , the ordinary Euclidean volume of the parallelotope
( n
)
X
ai vi : 0 ≤ ai ≤ 1
i=1
p
having edges vi is | det(vi · vj )|. The discriminant of an A-basis of B uses the A-valued
pairing hx, yi = TrB/A (xy) on B in place of the R-valued dot product on Rn and we just
drop the absolute value and the square root when we make the algebraic analogue.
How are the discriminants of two A-module bases for B related? Pick a second basis
e01 , . . . , e0n of B as an A-module. Then
e0i =
n
X
j=1
aij ej ,
DISCRIMINANTS AND RAMIFIED PRIMES
3
where aij ∈ A and the change of basis matrix (aij ) has determinant in A× . Then
!
n
n
X
X
0 0
TrB/A (ei ej ) = TrB/A
aik ek
aj` e`
k=1
=
n X
n
X
`=1
aik TrB/A (ek e` )aj` ,
k=1 `=1
so
(TrB/A (e0i e0j )) = (aij )(TrB/A (ei ej ))(aij )> .
Therefore
discA (e01 , . . . , e0n ) = (det(aij ))2 discA (e1 , . . . , en ).
We set
discA (B) = discA (e1 , . . . , en ) ∈ A
for any A-module basis {e1 , . . . , en } of B. It is well-defined up to a unit square. In particular,
the condition discA (B) = 0 is independent of the choice of basis.
Given a number field K, ramification of the prime p in K has been linked to the structure
of the ring OK /(p) in Section 1. Let’s look at the discriminant of this ring over Z/pZ. Letting
K have degree n over Q, the ring OK is a free rank-n Z-module, say
OK =
n
M
Zωi .
i=1
Reducing both sides modulo p,
OK /(p) =
n
M
(Z/pZ)ω i ,
i=1
so OK /(p) is a vector space over Z/pZ of dimension n. The discriminant of OK is discZ (OK ).
The next lemma says reduction modulo p commutes (in a suitable sense) with the formation
of discriminants.
Lemma 3.3. Choosing bases appropriately for OK and OK /(p),
discZ (OK ) mod p = discZ/pZ (OK /(p)).
Proof. Pick a Z-basis ω1 , . . . , ωn of OK . The reductions ω 1 , . . . , ω n in OK /(p) are a Z/pZbasis, so the multiplication matrix [mx ] for any x ∈ OK , with respect to the basis {ωi },
reduces modulo p to the multiplication matrix [mx ] for x on OK /(p) with respect to the
basis {ω i }. Therefore
Tr(OK /(p))/(Z/pZ) (x) = Tr(mx ) = Tr(mx ) mod p = TrOK /Z (x) mod p.
Thus, the mod p reduction of the matrix (TrOK /Z (ωi ωj )) is (Tr(OK /(p))/(Z/pZ) (ω i ω j )). Now
take determinants.
Lemma 3.4. Let A be a commutative ring and B1 and B2 be commutative ring extensions
of A which are each finite free A-modules. Then, choosing A-module bases appropriately,
discA (B1 × B2 ) = discA (B1 ) discA (B2 ).
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KEITH CONRAD
Proof. Pick A-module bases for B1 and B2 :
m
M
B1 =
Aei ,
B2 =
i=1
n
M
Afj .
j=1
As an A-module basis for B1 × B2 we will use the m + n elements e1 , . . . , em , f1 , . . . , fn .
Since ei fj = 0 in B1 × B2 , the matrix whose determinant is discA (B1 × B2 ) is a block
diagonal matrix
(Tr(B1 ×B2 )/A (ei ek ))
O
.
O
(Tr(B1 ×B2 )/A (fj f` ))
For any x ∈ B1 , multiplication by x on B1 × B2 kills the B2 component and acts on the
B1 -component in the way x multiplies on B1 , so a matrix for multiplication by x on B1 × B2
is a matrix whose upper left block is a matrix for multiplication by x on B1 and other blocks
are 0. Thus
Tr(B1 ×B2 )/A (x) = TrB1 /A (x) for x ∈ B1 .
Similarly, Tr(B1 ×B2 )/A (x) = TrB2 /A (x) for x ∈ B2 . Thus
(Tr(B1 ×B2 )/A (ei ek ))
O
(TrB1 /A (ei ek ))
O
=
,
O
(Tr(B1 ×B2 )/A (fj f` ))
O
(TrB2 /A (fj f` ))
and taking determinants gives
discA (B1 × B2 ) = discA (B1 ) discA (B2 ).
Now we prove Theorem 1.3.
Proof. We have p| discZ (OK ) if and only if discZ (OK ) ≡ 0 mod p. By Lemma 3.3
discZ (OK ) mod p = discZ/pZ (OK /(p)),
so p| discZ (OK ) if and only if discZ/pZ (OK /(p)) = 0 in Z/pZ.
In (1.2), each factor OK /pei i is a Z/pZ-vector space since p ∈ pei i . Using (1.2) and Lemma
3.4,
g
Y
discZ/pZ (OK /(p)) =
discZ/pZ (OK /pei i ).
i=1
Therefore we need to show for any prime number p and prime-power ideal pe such that
pe |(p) that discZ/pZ (OK /pe ) is 0 in Z/pZ if and only if e > 1. (Recall that the vanishing of
a discriminant is independent of the choice of basis.)
Suppose e > 1. Then any x ∈ p − pe is a nonzero nilpotent element in OK /pe . By linear
algebra over fields, such an x can be used as part of a Z/pZ-basis of OK /pe , say {x1 , . . . , xn }
with x = x1 . Writing the trace map Tr(OK /pe )/(Z/pZ) as Tr for short, the first column of the
matrix (Tr(xi xj )) contains the numbers Tr(xi x). These traces are all 0: xi x is nilpotent,
so the linear transformation mxi x on OK /pe is nilpotent and thus its eigenvalues all equal
zero. Since one column of the trace-pairing matrix (Tr(xi xj )) is all 0, discZ/pZ (OK /pe ) = 0.
Now suppose e = 1. Then OK /pe = OK /p is a finite field of characteristic p. We want
to prove discZ/pZ (OK /p) 6= 0. If this discriminant is 0, then (because OK /p is a field) the
trace function Tr : OK /p → Z/pZ is identically zero. However, from the theory of finite
fields, this trace function can be written as a polynomial function:
2
Tr(t) = t + tp + tp + · · · + tp
r−1
,
DISCRIMINANTS AND RAMIFIED PRIMES
5
where #(OK /p) = pr . Since the degree of this polynomial is less than the size of OK /p, this
function is not identically zero on OK /p. Therefore the discriminant of a finite extension of
Z/pZ does not equal zero.